Question

What is the solubility of PbCL2 in 0.4 M Kl given the solubility constant of PbCl2 is 7.1x 10^-9?
A. 12 x 10^-3 M
B. 3.5 x 10^-3 M
C. 4.4 x 10^-8 M
D. 1.9 x 10^-3 M

Answer 1

\[PbCl_{2} \rightarrow Pb^{2+} +2 Cl^{-}\]

Answer 2

I did not know there was even an equation

Answer 3

it's been a while but let me see. we know that PbCl we can represent this by [x]

\[PbCl_{2} \rightarrow Pb^{2+} +2 Cl^{-}\]

\[[Pb^{2}][2Cl^{-}]^{2} = K_{sp}\]

Answer 4

how do we get to the answer

Answer 5

This gives us an equation to find the concentration this is what i'm getting

i'll explain the whole thing to you

\[x^{3} = (7.1*10^{-9})\]

\[\sqrt[3]{ \frac{ 7.1*10^{-9} }{ } } = 1.9*10^{-3}\]

Answer 6

Just to add some information. You have to write the dissociation equation and balance it out to find out the ions that make up the ksp expression. Then you consider them as x like above and find the concentration for x . In this situation concentration of x is concentration of Pb. [Pb] does not equal to [2Cl]. Be careful of the COEFFICIENTS. Thus, [PbCl2]= [Pb]

Answer 7

Oh my gosh. people on here are so smart.

Answer 8

@Photon336 was that the explanation?

Answer 9

yeah @korosh23 is right. it's been a while since i've done this.

Answer 10

:)

Answer 11

Wait what was the answer. did i do it wrong then? @korosh23

Answer 12

@Photon336 provided you with the right answer.

Answer 13

thankyou

Answer 14

My pleasure.